Question: Let $h$ be a vector-valued function defined by $h(t)=(\cos(4\pi t),\sqrt{t+3})$. Find $h$ 's second derivative $h''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-16\pi^2\cos(4\pi t),-\dfrac{1}{4\sqrt{(t+3)^3}}\right)$ (Choice B) B $\left(-\cos(4\pi t),-\dfrac{3}{2\sqrt{(t+3)^3}}\right)$ (Choice C) C $-\cos(4\pi t)+\dfrac{1}{4\sqrt{(t+3)^3}}$ (Choice D) D $\left(-4\pi\sin(4\pi t),\dfrac{1}{2\sqrt{t+3}}\right)$
Answer: We are asked to find the second derivative of $h$. This means we need to differentiate $h$ twice. In other words, we differentiate $h$ once to find $h'$, and then differentiate $h'$ (which is a vector-valued function as well) to find $h''$. Recall that $h(t)=(\cos(4\pi t),\sqrt{t+3})$. Therefore, $h'(t)=\left(-4\pi\sin(4\pi t),\dfrac{1}{2\sqrt{t+3}}\right)$. Now let's differentiate $h'(t)=\left(-4\pi\sin(4\pi t),\dfrac{1}{2\sqrt{t+3}}\right)$ to find $h''$. $h''(t)=\left(-16\pi^2\cos(4\pi t),-\dfrac{1}{4\sqrt{(t+3)^3}}\right)$ In conclusion, $h''(t)=\left(-16\pi^2\cos(4\pi t),-\dfrac{1}{4\sqrt{(t+3)^3}}\right)$.